"\\mu= 60, \\sigma=10"
(a) P(X>70)
X is normally distributed with mean 60 and standard deviation 10.
"P(X>70)=1-P(X\\le70)"
"Z-value =\\frac{X-\\mu}{\\sigma}"
"=\\frac{70-60}{10}=1". From Z table or =NORM.DIST(70,60,10,TRUE),
"P(Z\\le1)= 0.841345"
Thus, "P(X>70)=1-0.841345=0.158655"
(b). "P(\\bar{X}>70)"
"\\bar{X}" Is normally distributed with mean 60 and variance "\\frac{10}{\\sqrt10}"
"P(\\bar{X}>70)=1-P(\\bar{X}\\le70)"
"Z-value =\\frac{\\bar{X}-\\mu}{\\frac{\\sigma}{\\sqrt{n}}}"
"=\\frac{70-60}{\\frac{10}{\\sqrt{10}}}=3.162278"
From Z table or =NORM.DIST(70,60,10/10^0.5,TRUE),"P(\\bar{X}\\le70)=" 0.999217
Thus, "P(\\bar{X}>70)=1-0.999217=0.000783"
(c). As the sample size increases, the sample mean tends to be closer to the population mean, and standard deviation decreases as the sample size increases. Part (a) has a sample size of 1 and the standard deviation is 10, while part (b) has a sample size of 10 leading to the standard deviation of"\\frac{10}{\\sqrt{10}}" which is smaller than the standard deviation in part (a). Therefore, part (a) has a greater probability since the distribution around X is wider than the distribution around "\\bar{X}" in part (b).
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